Efficient coding of side information in a lossless encoder

ABSTRACT

For “Super Audio CD” (SACD) the DSD signals are losslessly coded, using framing, prediction and entropy coding. Besides the efficiently encoded signals, a large number of parameters, i.e. the side-information, has to be stored on the SACD too. The smaller the storage capacity that is required for the side-information, the better the overall coding gain is. Therefore coding techniques are applied to the side-information too so as to compress the amount of data of the side information.

The invention relates to an apparatus for lossless encoding of a digitalinformation signal, for a lossless encoding method, to an apparatus fordecoding and to a record carrier.

For “Super Audio CD” (SACD) the DSD signals are losslessly coded, usingframing, prediction and entropy coding. Besides the efficiently encodedsignals, a large number of parameters, i.e. the side-information, has tobe stored on the SACD too. The smaller the storage capacity that isrequired for the side-information, the better the overall coding gainis. Therefore coding techniques are applied to the side-information too.A description of the lossless encoding of DSD signals is given in thepublication ‘Improved lossless coding of 1-bit audio signals’, by F.Bruekers et al, preprint 4563(I-6) presented at the 103^(rd) conventionof the AES, Sep. 26-29, 1997 in New York.

The invention aims at providing methods that can be used e.g. in SACD tosave on the number of bits that have to be used for storing theside-information. In the following description those methods will bepresented.

These and other aspects of the invention will be further explainedhereafter in the figure description, in which

FIG. 1 a shows a circuit diagram of a lossless encoder and FIG. 1 bshows a circuit diagram of a corresponding decoder, using linearprediction and arithmetic coding,

FIG. 2 shows subsequent frames of a multi channel information signal,

FIG. 3 shows the segmentation of time equivalent frames of the multichannel information signal, and

FIG. 4 shows the contents of a frame of the output signal of theencoding apparatus.

The process of lossless encoding and decoding, for the example of 1-bitoversampled audio signals, will be explained briefly hereafter by meansof FIG. 1, which shows an embodiment of the encoder apparatus in FIG. 1a and shows an embodiment of the decoder apparatus in FIG. 1 b.

The lossless coding in the apparatus of FIG. 1 a is performed onisolated parts (frames) of the audio signal. A typical length of such aframe is 37632 bits. The two possible bit-values of the input signal F,‘1’ and ‘0’, represent the sample values +1 and −1 respectively. Perframe, the set of coefficients for the prediction filter z⁻¹ .A(z),denoted by 4, is determined in a filter coefficient generator unit 12,by e.g. the autocorrelation method. The sign of the filter outputsignal, Z, determines the value of the predicted bit F_(p), whereas themagnitude of the filter output signal, Z, is an indication for theprobability that the prediction is correct. Upon quantizing the filteroutput signal Z in a quantizer 10, a predicted input signal F_(p) isobtained, which is ex-ored in a combining unit 2, resulting in aresidual signal E. A correct prediction, or F=F_(p), is equivalent toE=0 in the residual signal E. The content of the probability table,p(|.|), is designed per frame such that per possible value of Z, p₀ isthe probability that E=0. For small values of |Z| the probability for acorrect prediction is close to 0.5 and for large values of |Z| theprobability for a correct prediction is close to 1.0. Clearly theprobability for an incorrect prediction, F≠F_(p) or E=1, is p₁=1p₀.

The probability tables for the frames (or segments, to be describedlater) are determined by the unit 13. Using this probability table,supplied by the unit 13 to the unit 8, the unit 8 generates aprobability value P₀ in response to its input signal, which is thesignal Z.

The arithmetic encoder (AC Enc.) in the apparatus of FIG. 1 a, denotedby 6, codes the sequence of bits of E such that the code (D) requiresless bits. For this, the arithmetic coder uses the probability that bitn of signal E, E[n], has a particular value. The number of bits to codethe bit E[n]=0 is:d _(n)=−²log(p ₀)+ε(bits)  (Eq. 1)which is practically not more than 1 bit, since p₀≧½. The number of bitsto code the bit E[n]=1 is:d _(n)=−²log(p ₁)+ε=−² log(1−p ₀)+ε(bits)  (Eq. 2)which is not less than 1 bit. The ε in both equations represents thenon-optimal behavior of the arithmetic coder, but can be neglected inpractice.

A correct prediction (E[n]=0) results in less than 1 bit and anincorrect prediction (E[n]=1) results in more than 1 bit in the code(D). The probability table is designed such that on the average for thecomplete frame, the number of bits for code D is minimal.

Besides code D, also the coefficients of the prediction filter 4,generated by the coefficient generator unit 12, and the content of theprobability table, generated by the probability table determining unit13, have to be transmitted from encoder to decoder. To that purpose, theencoder apparatus comprises a multiplexer unit 14, which receives theoutput signal of the coder 6, as well as side information from thegenerator units 12 and 13. This side information comprises theprediction filter coefficients and the probability table. Themultiplexer unit 14 supplies the serial datastream of information to atransmission medium, such as a record carrier.

In the decoder apparatus of FIG. 1 b, exactly the inverse of the encoderprocess is implemented thus creating a lossless coding system. Thedemultiplexer unit 20 receives the serial datastream comprising the dataD and the side information. It retrieves the data D therefrom andsupplies the data D to an arithmetic decoder 22. The arithmetic decoder(AC Dec.) is provided with the identical probabilities as the arithmeticencoder was, to retrieve the correct values of signal E. Therefore thedemultiplexer unit retrieves the same prediction filter coefficients andprobability table as used the encoder from the serial datastreamreceived and supplies the prediction filter coefficients to theprediction filter 24 and the probability table to the probability valuegenerator unit 26.

The circuit constructions shown in FIG. 1 are meant forencoding/decoding a single serial datastream of information.Encoding/decoding a multi channel information signal, such as a multichannel digital audio signal, requires the processing described abovewith reference to FIG. 1 to be carried out in time multiplex by thecircuits of FIG. 1, or can be carried out in parallel by a plurality ofsuch circuits. Another solution can be found in international patentapplication IB 99/00313, which corresponds to U.S. Ser. No. 09/268,252(PHN 16.805).

It should be noted here that in accordance with the invention, theencoding apparatus may be devoid of the quantizer Q and the combiningunit 2. Reference is made to earlier patent publications discussingthis.

In SACD the 1-bit audio channels are chopped into frames of constantlength and per frame the optimal strategy for coding will be used.Frames can be decoded independently from neighbouring frames. Thereforewe can discuss the data structure within a single frame.

FIG. 2 shows time equivalent frames B of two channel signals, such asthe left and right hand signal component of a digital stereo audiosignal, indicated by . . . , B(l,m−1), B(l,m), B(l,m+1), . . . for theleft hand signal component and by . . . , B(r,m−1), B(r,m), B(r,m+1), .. . for the right hand signal component. The frames can be segmented, aswill be explained hereafter. If not segmented, the frames will beencoded in their entirety, with one set of filter coefficients and oneprobability table for the complete frame. If segmented, each segment ina frame can have its own set of filter coefficients and probabilitytable. Furthermore, the segmentation in a frame for the filtercoefficients need not be the same as for the probability tables. As anexample, FIG. 3 shows the two time equivalent frames B(l,m) and B(r,m)of the two channel signals being segmented. The frame B(l,m) has beensegmented into three segments fs(l,1), fs(l,2) and fs(l,3) in order tocarry out three different prediction filterings in the frame. It shouldhowever be noted that the filterings in two segments, such as thesegments fs(l,1) and fs(l,3) can be the same. The frame B(l,m) hasfurther been segmented into two segments ps(l,1) and ps(l,2) in order tohave two different probability tables for those segments.

The frame B(r,m) has been segmented into three segments fs(r,1), fs(r,2)and fs(r,3) in order to carry out three different prediction filteringsin the frame. It should however again be noted that the filterings intwo segments, such as the segments fs(r,1) and fs(r,3) can be the same.The frame B(r,m) has further been segmented into four segments ps(r,1),ps(r,2), ps(r,3) and ps(r,4) in order to have four different probabilitytables for those segments. Again, it should be noted that some of thesegments can have the same probability table.

The decision to have the same probability table for different segmentscan be taken on beforehand by a user of the apparatus, after havingcarried out a signal analysis on the signals in the segments. Or theapparatus may be capable of carrying out this signal analysis and decidein response thereto. In some situations, a signal analysis carried outon two segments may result in probability tables that differ onlyslightly. In such situation, it can be decided to have one and the sameprobability table for both segments. This one probability table could beequal to one of the two probability tables established for the twosegments, or could be an averaged version of both tables. An equivalentreasoning is valid for the sets of filter coefficients in the varioussegments.

To summarize: in order to encode a small portion of audio in an audiochannel signal, the coding algorithm in SACD requires both a predictionfilter (the filter) and a probability table (the table). For improvingthe coding gain it can be efficient to use different filters indifferent channels. But also within the same channel is can bebeneficial to use different filters. That is why the concept ofsegmentation is introduced. A channel is partitioned into segments andin a segment a particular filter is used. Several segments, also fromother channels, may use the same or a different filter. Besides storageof the filters that are used, also information about the segments(segmentation) and information about what filter is used in what segment(mapping) have to be stored.

For the tables, the same idea is applicable, however the segmentationand mapping may be different from the segmentation and mapping for thefilters. In case of equal segmentation for both filter and table this isindicated. The same idea is used for the mapping. If the segmentationfor the filters is equal for all channels this is indicate too. The sameidea is used for the mapping.

First, a description will be given of the contents of a frame of atransmission signal comprising the encoded channel signals and thecorresponding side information. FIG. 4 shows a schematic drawing of theframe. Apart from synchronization information (not shown), the framecomprises two words w₁ and w₂, followed by segmentation information onthe prediction filters. Next a word w₃ is present followed bysegmentation information on the probability tables. Next follow twowords w₄ and w₅, followed by mapping information on the predictionfilters. Next, follows a word w₆, followed by mapping information on theprobability tables. Next follow the filter coefficients and theprobability tables, as supplied by the generator units 12 and 13,respectively. The frame ends with the data D, supplied by the arithmeticencoder 6.

The word w₁ is in this example one bit long and can have the value ‘0’or ‘1’, and indicates whether the segment information for the filtercoefficients and the probability tables are the same (‘1’), or not(‘0’). The word w₄ is in this example one bit long and can have thevalue ‘0’ or ‘1’, and indicates whether the mapping information for thefilter coefficients and the probability tables are the same (‘1’), ornot (‘0’). The word w₂, again one bit long, can have the value ‘0’ or‘1’, and indicates whether the channel signals have the samesegmentation information for the prediction filter coefficients (‘1’),or not (‘0’). The word w₃ (one bit long) can have the value ‘0’ or ‘1’,and indicates whether the channel signals have the same segmentationinformation for the probability tables (‘1’), or not (‘0’). The word w₅can have the value ‘0’ or ‘1’, and indicates whether the channel signalshave the same mapping information for the prediction filter coefficients(‘1’), or not (‘0’). The word w₆ can have the value ‘0’, and indicateswhether the channel signals have the same mapping information for theprobability tables (‘1’), or not (‘0’).

First, the representation of the total number of segments S in a framewill be described.

To code a number, e.g. the total number of segments in a frame in aparticular channel signal, a kind of run-length coding is applied. It isimportant that the code is short for small values of S. Since the numberof segments in a channel S≧1, S=0 needs not to be coded. In SACD thefollowing codes are used.

TABLE 1 S code(S) 1 1 2 01 3 001 4 0001 s 0^((s−1))1

Remark: Here the “1” is used as delimiter. It is clear that in generalthe role of the “0” and “1” can be interchanged. The basic idea of thedelimiter is that a certain sequence is violated; the sequence of “0's”is violated by a “1”. An alternative is e.g. to “inverse” the nextsymbol and “no inversion” is used as a delimiter. In this way longconstant sequences are avoided. An example of inverting sequences thatstart with an “1” is (not used in SACD):

S code(S) 1 0 2 11 3 100 4 1011 5 10100 6 101011

Second, the representation of the segment sizes will be described. Thelength of a segment will be expressed in number of bytes of the channelsignal. The B bytes in a frame of a channel signal are partitioned intoS segments. For the first S−1 segments the number of bytes of eachsegment has to be specified. For the S^(th) segment the number of bytesis specified implicitly, it is the remaining number of bytes in thechannel. The number of bytes in segment i, equals

-   B_(i) so the number of bytes in the last segment is:

$B_{s - 1} = {B - {\sum\limits_{i = 0}^{s - 2}B_{i}}}$Since the number of bytes in the first S−1 segments are multiples of Rthe resolution R≧1, we define:

-   B_(i)=b_(i)R and consequently:

$B_{s - 1} = {B - {\sum\limits_{i = 0}^{s - 2}{b_{i}R}}}$The S−1 values of b_(i) are stored and R is stored in a channel only ifS>1 and when it is not stored already for another channel.The number of bits required to store b_(i) depends on its possiblevalues.0≦b_(i)≦b_(i,max) with e.g.

$b_{i,\max} = {\left\lfloor \frac{B}{R} \right\rfloor - {\sum\limits_{j = 0}^{i - 1}b_{j}}}$so the required number of bits to store b_(i) is:#bits(b _(i))=└²log(b _(i,max))┘+1

This has as advantage that the required number of bits for the segmentlength may decrease for segments at the end of the frame. Ifrestrictions are imposed on e.g. minimal length of a segment thecalculation of the number of bits may be adapted accordingly. The numberof bits to store the resolution R is: #bits(R)

Third, the representation of the segmentation information in the serialdatastream will be described. Use will be made of the representationsgiven above under table 1. This will be illustrated by some examples.

In order to distinguish between filters and probability tables, thesubscripts ƒ and t are used. To distinguish between segments indifferent channels the double argument is used: (channel number, segmentnumber).

Next follows a first example. For a 2-channel case, we have differentsegmentations for filters and probability tables, and the segmentationis different for both channels. The following table shows the parametersin the stream.

TABLE 2 Value #bits comment (w₁ =) 0 1 segmentation information for thefilters is different from the segmentation for the probability tablesfilter segmentation (w₂ =) 0 1 channels have own filter segmentationinformation filter segmentation in channel 0 (y₁ =) 0 1 first bit ofcode(S_(f)(0)) indicating that S_(f)(0) ≧ 2 R_(f) #bits(R_(f))resolution for filters b_(f)(0,0) #bits(b_(f)(0,0)) first segment inchannel 0 has length R_(f) b_(f)(0,0) bytes (y₂ =) 1 1 last bit ofcode(S_(f)(0)) indicating that S_(f)(0) = 2 filter segmentation inchannel 1 (y₁ =) 0 1 first bit of code(S_(f)(1)) indicating thatS_(f)(1) ≧ 2 b_(f)(1,0) #bits(b_(f)(1,0)) first segment in channel 1 haslength R_(f) b_(f)(1,0) bytes (y₂ =) 0 1 second bit of code(S_(f)(1))indicating that S_(f)(1) ≧ 3 b_(f)(1,1) #bits(b_(f)(1,1)) second segmentin channel 1 has length R_(f) b_(f)(1,1) bytes (y₃ =) 1 1 last bit ofcode(S_(f)(1)) indicating that S_(f)(1) = 3 probability tablesegmentation (w₃ =) 0 1 channels have own table segmentationspecification probability table segmentation in channel 0 (y₁ =) 1 1last bit of code(S_(t)(0)) indicating that S_(t)(0) = 1 probabilitytable segmentation in channel 1 (y₁ =) 0 1 first bit of code(S_(t)(1))indicating that S_(t)(1) ≧ 2 R_(t) #bits(R_(t)) resolution for tablesb_(t)(1,0) #bits(b_(t)(1,0)) first segment in channel 1 has length R_(t)b_(t)(1,0) bytes (y₂ =) 0 1 second bit of code(S_(t)(1)) indicating thatS_(t)(1) ≧ 3 b_(t)(1,1) #bits(b_(t)(1,1)) second segment in channel 1has length R_(t) b_(t)(1,1) bytes (y₃ =) 1 1 last bit of code(S_(t)(1))indicating that S_(t)(1) = 3

In the above table 2, the first combination (y₁,y₂) equal to (0,1) isthe codeword code(S) in table 1 above, and indicates that in the channelsignal numbered 0 the frame is divided into two segments for the purposeof prediction filtering. Further, the combination (y₁,y₂,y₃) equal to.(0,0,1) is the codeword code(S) in table 1 above, and indicates that inthe channel signal numbered 1 the frame is divided into three segmentsfor the purpose of prediction filtering. Next, we find a combination(ye) equal to (1), which is the first codeword in table 1, indicatingthat the channel signal numbered 0, the frame is not divided for theprobability table. Finally, we find a combination (y₁,y₂,y₃) equal to(0,0,1), which indicates that the frame of the second channel signal isdivided into three segments, each with a corresponding probabilitytable.

Next, follows another example for a 5-channel case. It is assumed thatfor this 5-channel case, we have equal segmentation for filters andtables, and the segmentation is equal for all channels.

Value #bits comment (w₁ =) 1 1 the segmentation information for theprediction filters and probability tables is the same filtersegmentation (w₂ =) 1 1 channels have equal filter segmentation informa-tion filter segmentation in channel 0 (y_(t) =) 0 1 first bit ofcode(S_(f)0)) indicating that S_(f)(0) ≧ 2 R_(f) #bits(R_(f)) resolutionfor filters b_(f)(0,0) #bits(b_(f)0,0)) first segment in channel 0 haslength R_(f) b_(f)(0,0) bytes (y₂ =) 0 1 second bit of code(S_(f)(0))indicating that S_(f)(0) ≧ 3 b_(f)(0,1) #bits(b_(f)0,1)) second segmentin channel 0 has length R_(f) b_(f)(0,1) bytes (y₃ =) 1 1 last bit ofcode(S_(f)(0)) indicating that S_(f)(0) = 3 filter segmentation inchannel c b_(f)(c,0) = b_(f)(0,0) and b_(f)(c,1) = b_(f)(0,1) for 1 ≦ c< 5 probability table segmentation in channel c b_(t)(c,0) = b_(f)(0,0)and b_(t)(c,1) = b_(f)(0,1) for 0 ≦ c < 5

Remark: The single bits of code(S) interleaved in de segmentationinformation can be interpreted as “another segment will be specified” incase of “0” or “no more segments will be specified” in case of “1”.

Next, mapping will be described.

For each of the segments, all segments of all channels are consideredtogether, it has to be specified which filter or table is used. Thesegments are ordered; first the segments of channel 0 followed by thesegments of channel 1 and so on.

The filter or table number for segment s, N(s) is defined as:

$\left\{ \begin{matrix}{{N(0)} = 0} \\{0 \leq {N(s)} \leq {N_{\max}(s)}}\end{matrix}\quad \right.$with N_(max) (s), the maximum allowed number for a given segment,defined as:N _(max)(s)=1+max(N(i)) with 0≦i<sThe required number of bits to store N(s) equals:#bits(N(s))=└²log(N _(max)(s))┘+1

The number of bits that is required to store a filter or table numberaccording to this method depends on the set of numbers that already hasbeen assigned.

If the tables use the same mapping as the filters, which is not alwayspossible, this is indicated. Also when all channels use the same mappingthis is indicated.

With two examples the idea will be illustrated.

EXAMPLE 3

Assume that in total we have 7 segments (0 through 6), some segments usethe same filter and some use a unique filter. Furthermore it is assumedthat the tables use the same mapping specification as the filters.

Channel Segment Filter Possible filter number number number numbers#bits 0 0 0 — 0 1 1 0 0 or 1 1 1 2 1 0 or 1 1 1 3 2 0, 1 or 2 2 1 4 3 0,1, 2 or 3 2 2 5 3 0, 1, 2, 3 or 4 3 3 6 1 0, 1, 2, 3 or 4 3 Total #bits12 

Segment number 0 uses filter number 0 per definition, so no bits areneeded for this specification. Segment number 1 may use an earlierassigned filter (0) or the next higher not yet assigned filter (1), so 1bit is needed for this specification. Segment number 1 uses filternumber 0 in this example. Segment number 2 may use an earlier assignedfilter (0) or the next higher not yet assigned filter (1), so 1 bit isneeded for this specification. Segment number 2 uses filter number 1 inthis example.

Segment number 3 may use an earlier assigned filter (0 or 1 ) or thenext higher not yet assigned filter (2), so 2 bits are needed for thisspecification. Segment number 3 uses filter number 2 in this example.

Segment number 4 may use an earlier assigned filter (0, 1 or 2) or thenext higher not yet assigned filter (3), so 2 bits are needed for thisspecification. Segment number 4 uses filter number 3 in this example.Segment number 5 may use an earlier assigned filter (0, 1, 2 or 3) orthe next higher not yet assigned filter (4), so 3 bits are needed forthis specification. Segment number 5 uses filter number 3 in thisexample.

Segment number 6 may use an earlier assigned filter (0, 1, 2 or 3) orthe next higher not yet assigned filter (4), so 3 bits are needed forthis specification. Segment number 6 uses filter number 1 in thisexample.

In total 12 bits are required to store the mapping. The total number ofsegments (7 segments in this example) is known at this point in thestream.

Value #bits comment (w₄ =) 1 1 probability tables have same mappinginformation as prediction filters prediction filter mapping (w₅ =) 0 1channels have own filter, segmentation information 0 filter number forsegment 0 is 0 per definition N_(f)(1) #bits(N_(f)(1)) filter number forsegment 1 N_(f)(2) #bits(N_(f)(2)) filter number for segment 2 N_(f)(3)#bits(N_(f)(3)) filter number for segment 3 N_(f)(4) #bits(N_(f)(4))filter number for segment 4 N_(f)(5) #bits(N_(f)(5)) filter number forsegment 5 N_(f)(6) #bits(N_(f)(6)) filter number for segment 6probability table mapping N_(t)(i) = N_(f)(i) for 0 ≦ i < 7Another example. Assume that in total we have 6 channels each with 1segment and each segment uses the same prediction filter and the sameprobability table.

Value #bits comment (w₄ =) 1 1 probability tables have same mappinginformation as prediction filters prediction filter mapping (w₅ =) 1 1channels have own filter mapping specification 0 filter number forsegment 0 is 0 per definition prediction filter mapping for segment iN_(f)(i) = 0 for 1 ≦ i < 6 probability table mapping for segment iN_(t)(i) = N_(f)(i) for 0 ≦ i < 6In total 2 bits are required to store the complete mapping.

Remark: A reason to give the indication that a following specificationis also used for other application (e.g. for tables the samesegmentation is used as for the filters) is that this simplifies thedecoder.

Whilst the invention has been described with reference to preferredembodiments thereof, it is to be understood that these are notlimitative examples. Thus, various modifications may become apparent tothose skilled in the art without departing from the scope of theinvention as defined by the claims. As an example, the invention couldalso have been incorporated in an embodiment in which time equivalentsignal blocks are encoded, without making use of segmentation. In suchembodiment, the serial datastream obtained, like the datastream of FIG.4, will be devoid of the segment information described there for thefilters and the probability tables, as well as some of the indicatorwords, such as the indicator words w₁, w₂ and w₃. Further, the inventionlies in each and every novel feature and combination of features.

1. A method for encoding of a n-channel digital audio signal, where n isan integer larger than 1, said method the steps of: encoding eachchannel signals of the n-channel digital audio signal so as to obtain anencoded channel signal for each of said channel signals in response toprobability values for each of said channel signals; carrying out aprediction filtering on each of said channel signals in response to aset of prediction filter coefficients for each of said channel signalsso as to obtain a prediction filtered channel signal from each of saidchannel signals; generating a set of prediction filter coefficients foreach of said channel signals; generating probability values for each ofsaid channel signals in response to a probability table for each of saidchannel signals and the corresponding prediction filtered channel signalfor each of said channel signals; generating the probability tables foreach of said channel signals; generating first mapping information for aplurality of m sets of prediction filter coefficients, where m is aninteger for which holds 1≦m≦n, said first mapping information and m setsof prediction filter coefficients being representative of said n sets ofprediction filter coefficients for said n channels; generating secondmapping information for a plurality of p probability tables, where p isan integer for which holds 1≦p≦n, said second mapping information and pprobability tables being representative of said n probability tables forsaid n channels; combining said encoded channel signal for each of saidchannel signals, said first and second mapping informations saidplurality of m sets of prediction filter coefficients and said pluralityof p probability tables into a composite information signal; andoutputting said composite information signal as the encoded n-channeldigital audio signal.
 2. A method for encoding a n-channel digital audiosignal, where n is an integer larger than 1, said method comprising thesteps of: encoding time equivalent signal blocks of each channel signalsof the n-channel digital audio signal by dividing the time equivalentsignal blocks into M segments, and encoding the signal portions of thechannel signals in all M segments in said time equivalent signal blocks,so as to obtain an encoded signal portion for each of said signalportions in said M segments in response to probability values for eachof said signal portions, where M=Σ_(i=0) ^(i=n−1)sp_(i), and sp_(i) isthe number of segments in the time equivalent signal block of the i-thchannel signal; generating probability values for each of said M signalportions in response to a probability table for each of said M signalportions; generating the probability tables for each of said M signalportions; converting the information about the length and locations ofthe M segments in the n channel signals into first segment information,and generating first mapping information for a plurality of mprobability tables, where m is an integer for which holds 1≦m≦M saidfirst mapping information and said m probability tables beingrepresentative for said M probability tables; combining said encodedsignal portion for each of said signal portions, said first segmentinformation, said first mapping information signal and said plurality ofm probability tables into a composite information signal; and outputtingsaid composite information signal as the encoded n-channel digital audiosignal.
 3. The method as claimed in claim 2, wherein said method furthercomprises the steps of: predicting time equivalent signal blocks foreach of said channel signals by dividing the time equivalent signalblocks into segments, and prediction filtering the signal portions ofthe channel signals in all P segments in said time equivalent signalblocks, so as to obtain a prediction filtered signal portion for each ofsaid P signal portions in response to a set of prediction filtercoefficients for each of said signal portions, where P=Σ_(i=0) ^(i=n−1)sf_(i) and sf_(i) is the number of segments in the time equivalentsignal block of the i-th channel signal; generating a set of predictionfilter coefficients for each of said P signal portions; converting theinformation about the length and locations of the P segments in the nchannel signals into second segment information, and generating secondmapping information and a plurality of p sets of prediction filtercoefficients, where p is an integer for which holds 1≦p≦p, said secondmapping information and said p sets of prediction filter coefficientsbeing representative of said P sets of prediction filter coefficients;and combining said second segment information, said second mappinginformation signal and said plurality of p sets of prediction filtercoefficients into said composite information signal.
 4. The method asclaimed in claim 3, wherein said method further comprises the steps of:generating a first indicator word (w₁) of a first value, indicating thatthe segmentation of the time equivalent signal blocks for theprobability tables is different from the segmentation of the timeequivalent signal blocks for the sets of prediction filter coefficientsand of a second value indicating that the segmentation of the timeequivalent signal blocks for the probability tables is the same as forthe prediction filter coefficients; and supplying only one of the firstor the second segment information in the latter case, wherein thecombining step combines the first indicator word and the only one of thefirst segment information or the second segment information into saidcomposite information signal, in the case that the first indicator wordhas the second value.
 5. The method of claim 4, wherein said only one ofthe first or second segment information is generated if the firstindicator word has the second value.
 6. The method as claimed in claim3, wherein said method further comprises the step of: generating asecond indicator word (w₂) of a third value indicating that the timeequivalent signal blocks all have the same segmentation for the sets ofprediction filter coefficients, and of a fourth value indicating thatthe time equivalent signal blocks have each a different segmentation forthe sets of prediction filter coefficients, wherein the converting stepgenerates second segment information for only one time equivalent signalblock in the case that the second indicator word has the third value,and generates second segment information for each of the time equivalentsignal blocks in the case that the second indicator word has the fourthvalue, and wherein the combining combines the second indicator word intosaid composite information signal.
 7. The method as claimed in claim 2,wherein the method further comprises the step of: generating a thirdindicator word (w₃) of a fifth value indicating that the time equivalentsignal blocks all have the same segmentation for the probability tables,and of a sixth value indicating that the time equivalent signal blockshave each a different segmentation for the probability tables, whereinthe converting step generates first segment information for only onetime equivalent signal block in the case that the third indicator wordhas the fifth value, and generates first segment information for each ofthe time equivalent signal blocks in the case that the third indicatorword has the sixth value, and wherein the combining step combines thethird indicator word into said composite information signal.
 8. Themethod as claimed in claim 3, wherein the method further comprises thestep of: generating a fourth indicator word (w₄) of a seventh value,indicating that the mapping information for the probability tables isdifferent from the mapping information for the prediction filtercoefficients, and of an eighth value indicating that the mappinginformation for the probability tables is the same as for the predictionfilter coefficients, and for supplying the first or the second mappinginformation only in the latter case, wherein the combining means stepcombines the fourth indicator word and the first mapping information orthe second mapping information only into said composite informationsignal, in the case that the fourth indicator word has the eighth value.9. The method as claimed in claim 3, wherein the method furthercomprises the step of: generating a fifth indicator word (w₅) of a ninthvalue indicating that the time equivalent signal blocks all have thesame mapping information for the sets of prediction filter coefficients,and of a tenth value indicating that the time equivalent signal blockshave each a different mapping information for the sets of predictionfilter coefficients, wherein the converting step generates secondmapping information for only one time equivalent signal block in thecase that the fifth indicator word has the ninth value, and generatessecond mapping information for each of the time equivalent signal blocksin the case that the fifth indicator word has the tenth value, andwherein the combining step combines the fifth indicator word into saidcomposite information signal.
 10. The method as claimed in claim 2,wherein said method further comprises the step of: convertinginformation concerning the number of segments in a time equivalentsignal block of a channel signal into a number code, wherein thecombining step combines the number code into said composite informationsignal.
 11. The method as claimed in claim 10, wherein said number codesatisfies the following table: S code(S) 1 1 2 01 3 001 4 0001 s0^((s−1))1

where S is the number of segments in a time equivalent signal block of achannel signal.
 12. The method as claimed in claim 3, wherein the firstset of prediction filter coefficients is allocated to the first of saidP segments, said second mapping information being devoid of mappinginformation for mapping said first set of prediction filter coefficientsto said first segment of said P segments, (a) the first bit in saidsecond mapping information indicating whether the set of predictionfilter coefficients for the second segment is the first set ofprediction filter coefficients or a second set of prediction filtercoefficients, (b1) if the first set of prediction filter coefficients isalso the set of filter coefficients for the second segment, then thesecond bit in said second mapping information indicating whether the setof prediction filter coefficients for the third segment is the first setof prediction filter coefficients or the second set of prediction filtercoefficients, (b2) if the second set of prediction filter coefficientsis the set of filter coefficients for the second segment, then the nexttwo bits in the second mapping information indicating whether the set ofprediction filter coefficients for the third segment is the first, thesecond or the third set of prediction filter coefficients, (c1) if thefirst set of prediction filter coefficients is the set of filtercoefficients for the second and third segment, then the third bit ofsaid second mapping information indicates whether the set of predictionfilter coefficients for the fourth segment is the first or the secondset of prediction filter coefficients, (c2) if the first set ofprediction filter coefficients is the set of filter coefficients for thesecond segment and the second set of prediction filter coefficients isthe set of filter coefficients for the third segment, then the third andfourth bit in said second mapping information indicating whether the setof prediction filter coefficients for the fourth segment is the first,the second or the third set of prediction filter coefficients, (c3) ifthe second set of prediction filter coefficients is the set of filtercoefficients for the second segment, and the first or the second set offilter coefficients is the set of filter coefficients for the thirdsegment, then the fourth and fifth bit in the second mapping informationindicating whether the set of prediction filter coefficients for thefourth segment is the first, second or the third set of predictionfilter coefficients, (c4) if the second set of prediction filtercoefficients is the set of filter coefficients for the second segment,and the third set of filter coefficients is the set of prediction filtercoefficients for the third segment, then the fourth and fifth bit in thesecond mapping information indicating whether the set of predictionfilter coefficients for the fourth segment is the first, second, thirdor the fourth set of filter coefficients.
 13. The method as claimed inclaim 2, wherein the first probability table is allocated to the firstof said M segments, said first mapping information being devoid ofmapping information for mapping said first probability table to saidfirst segment of said M segments, (a) the first bit in said firstmapping information indicating whether the probability table for thesecond segment is the first probability table or a second probabilitytable, (b1) if the first probability table is also the probability tablefor the second segment, then the second bit in said first mappinginformation indicating whether the probability table for the thirdsegment is the first probability table or the second probability table,(b2) if the second probability table is the probability table for thesecond segment, then the next two bits in the first mapping informationindicating whether the probability table for the third segment is thefirst, the second or the third probability table, (c1) if the firstprobability table is the probability table for the second and thirdsegment, then the third bit of said first mapping information indicateswhether the probability table for the fourth segment is the first or thesecond probability table, (c2) if the first probability table is theprobability table for the second segment and the second probabilitytable is the probability table for the third segment, then the third andfourth bit in said first mapping information indicating whether theprobability table for the fourth segment is the first, the second or thethird probability table, (c3) if the second probability table is theprobability table for the second segment, and the first or the secondprobability table is the probability table for the third segment, thenthe fourth and fifth bit in the first mapping information indicatingwhether the probability table for the fourth segment is the first,second or the third probability table, (c4) if the second probabilitytable is the probability table for the second segment, and the thirdprobability table is the probability table for the third segment, thenthe fourth and fifth bit in the first mapping information indicatingwhether the probability table or the fourth segment is the first,second, third or the fourth probability table.
 14. A method for decodinga composite information signal comprising encoded data of a n-channeldigital audio signal and side information having a relationship with theencoded data, comprising the steps of: retrieving said encoded data andside information from said composite information signal; decoding theencoded data so as to obtain n channel signals in response to a set ofprobability values for each of said channel signals; carrying out aprediction filtering on each of said channel signals in response to nsets of prediction filter coefficients, one set for each of said channelsignals, so as to obtain a prediction filtered channel signal from eachof said channel signals, said sets of prediction filter coefficientsbeing derived from said side information; generating n sets ofprobability values, one for each of the channel signals in response to acorresponding prediction filtered channel signal and correspondingprobability table, said n probability tables, one for each of thechannel signals, being derived from said side information; retrievingfirst and second mapping information, a plurality of m sets ofprediction filter coefficients and a plurality of p probability tablesfrom said side information; reconverting said first mapping informationand said m sets of prediction filter coefficients into n sets ofprediction filter coefficients, one set for each of said channelsignals, where m is an integer for which holds 1≦m≦n; reconverting saidsecond mapping information and said p probability tables into nprobability tables, one set for each of said channel signals, where p isan integer for which holds 1≦p≦n; and outputting said n channel signalsas constituting said n-channel digital audio signal.
 15. A method fordecoding a composite information signal comprising encoded data of an-channel digital audio signal and side information having arelationship with the encoded data, where n is an integer larger than 1,comprising the steps of: retrieving said encoded data and sideinformation from said composite information signal; decoding saidencoded data into M signal portions in response to corresponding sets ofprobability values, one for each of said M signal portions, where$M = {\sum\limits_{i = 0}^{i = {n - 1}}{sp}_{i}}$ and sp_(i) is thenumber of segments in the time equivalent signal block of the i-thchannel signal; generating M sets of probability values, one for each ofthe M signal portions in response to a corresponding probability table,said M probability tables, one for each of the signal portions, beingderived from said side information; retrieving first segment informationand first mapping information and a plurality of m probability tablesfrom said side information, where m is an integer for which holds 1≦m≦M;reconverting said first mapping information and m probability tablesinto M probability tables, one for each of said signal portions;reconverting said first segment information into information about thelength and locations of the M segments in the n channel signals so as toobtain time equivalent signal blocks in said n channel signals; andoutputting said M signal portions as constituting said n-channel digitalaudio signal.
 16. The method as claimed in claim 15, wherein said methodfurther comprises the step of: outputting the time equivalent signalblocks of said n channel signals.
 17. The method as claimed in claim 15,wherein said method further comprises the steps of: carrying out aprediction filtering on said time equivalent signal blocks of each ofsaid channel signals by dividing the time equivalent signal blocks intosegments; prediction filtering the signal portions of the channelsignals in all P segments in said time equivalent signal blocks and forall n channel signals, so as to obtain a prediction filtered signalportion for each of said P signal portions in response to a set ofprediction filter coefficients for each of said signal portions, where$P = {\sum\limits_{i = 0}^{i = {n - 1}}{sf}_{i}}$ and sf_(i) is thenumber of segments in the time equivalent signal block of the i-thchannel signal; retrieving second segment information, second mappinginformation and p sets of prediction filter coefficients from said sideinformation, where p is an integer for which holds 1≦p≦P; reconvertingthe second segment information into information about the length andlocations of the P segments in the n channel signals; and reconvertingthe p sets of prediction filter coefficients into P sets of predictionfilter coefficients, one for each of said P signal portions, using saidsecond mapping information.
 18. The method as claimed in claim 17,wherein said method further comprises the steps of: retrieving a firstindicator word (w₁) from said side information, said first indicatorword, when being of a first value, indicating that the segmentation ofthe time equivalent signal blocks for the probability tables isdifferent from the segmentation of the time equivalent signal blocks forthe prediction filter coefficients, and when being of a second value,indicating that the segmentation of the time equivalent signal blocksfor the probability tables is the same as for the prediction filtercoefficients; and retrieving one segment information only from the sideinformation in the latter case, the reconverting means further beingadapted to copy the said segment information so as to obtain the firstand second segment information, in the latter case.
 19. The method asclaimed in claim 17, wherein said method further comprises the step of:retrieving a second indicator word (w₂) from said side information, saidsecond indicator word, when being of a third value, indicating that thetime equivalent signal blocks all have the same segmentation for theprediction filter coefficients and, when being of a fourth value,indicating that the time equivalent signal blocks have each a differentsegmentation for the prediction filter coefficients, said retrievingstep retrieving second segment information for only one time equivalentsignal block from the side information in the case that the secondindicator word has the third value, and retrieving second segmentinformation for each of the time equivalent signal blocks in the casethat the second indicator word has the fourth value, wherein thereconverting step copies the second segment information n−1 times so asto obtain the P segments of the time equivalent signal blocks of all nchannel signals, in the case that the second indicator word has thethird value.
 20. The method as claimed in claim 17, wherein theretrieving step retrieves a third indicator word (w₃) from said sideinformation, said third indicator word, when being of a fifth value,indicating that the time equivalent signal blocks all have the samesegmentation for the probability tables, and when being of a sixthvalue, indicating that the time equivalent signal blocks have each adifferent segmentation for the probability tables, said retrieving stepretrieving first segment information for only one time equivalent signalblock in the case that the third indicator word has the fifth value, andretrieving first segment information for each of the time equivalentsignal blocks in the case that the third indicator word has the sixthvalue, wherein the reconverting step copies the first segmentinformation for said one time equivalent signal block n−1 times so as toobtain the M segments of the time equivalent signal blocks of all the nchannel signals, in the case that the third indicator word has the fifthvalue.
 21. The method as claimed in claim 17, wherein said methodfurther comprises the step of: retrieving a fourth indicator word (w₄)from said side information, said fourth indicator word, when being of aseventh value, indicating that the mapping information for theprobability tables is different from the mapping information for thesets of prediction filter coefficients, and when being of an eighthvalue, indicating that the mapping information for the probabilitytables is the same as for the prediction filter coefficients, saidretrieving step retrieving only one mapping information from the sideinformation in case the fourth indicator word has the eighth value,wherein the reconverting step copies the mapping information retrievedin the case that the fourth indicator word has the eighth value.
 22. Themethod as claimed in claim 17, wherein the method further comprises thestep of: retrieving a fifth indicator word (w₅) from said sideinformation, said fifth indicator word, when being of a ninth value,indicating that the time equivalent signal blocks all have the samemapping information for the prediction filter coefficients, and whenbeing of a tenth value, indicating that the time equivalent signalblocks have each a different mapping information for the predictionfilter coefficients, said retrieving step retrieving second mappinginformation for only one time equivalent signal block in the case thatthe fifth indicator word has the ninth value, and retrieves secondmapping information for each of the time equivalent signal blocks in thecase that the fifth indicator word has the tenth value.
 23. The methodas claimed in claim 15, wherein said method further comprises the stepof: converting information to retrieve a number code for a timeequivalent signal block from said side information, said number coderepresenting the number of segments in said time equivalent signalblock.
 24. The method as claimed in claim 23, wherein said number codesatisfies the following table: S code(S) 1 1 2 01 3 001 4 0001 s0^((s−1))1

where S is the number of segments in a time equivalent signal block of achannel signal.